Problem: Solve for $z$, $ \dfrac{4z - 2}{3z + 5} = \dfrac{1}{7} $
Solution: Multiply both sides of the equation by $3z + 5$ $ 4z - 2 = \dfrac{3z + 5}{7} $ Multiply both sides of the equation by $7$ $ 7(4z - 2) = 3z + 5 $ $28z - 14 = 3z + 5$ $25z - 14 = 5$ $25z = 19$ $z = \dfrac{19}{25}$